Note that the linear independence of the vi’s is essential in insuring the existence of a linear transformation f. Otherwise, suppose 0=∑di⁢vi where d1≠0. Pick wi’s so that they are linearly independent. Then 0=f⁢(∑di⁢vi)=∑di⁢f⁢(vi)=∑di⁢wi, contradicting the linear independence of the wi’s.

Proof.

First, assume that R is a dense ring of linear transformations of V. Recall that the compact-open topology on EndD⁡(V) has subbasis of the form B⁢(K,U):={f∣f⁢(K)⊆U}, where U is open and K is compact in V. Since V is discrete, K is finite. Now, pick a point g∈EndD⁡(V) and let

B=⋃α∈I⋂i=1n⁢(α)B⁢(Ki⁢α,Ui⁢α)

be a neighborhood of g, I some index set. Then for some α∈I, g∈⋂B⁢(Ki⁢α,Ui⁢α). This means that g⁢(Ki⁢α)⊆Ui⁢α for all i=1,…,n⁢(α). Since each Ki⁢α is finite, so is K:=⋃Ki⁢α. After some re-indexing, let {v1,…,vn} be a maximal linearly independent subset of K. Set wj=g⁢(vj), j=1,…,n. By assumption, there is an f∈R such that f⁢(vj)=wj, for all j. For any v∈K, v is a linear combination of the vj’s: v=∑dj⁢vj, dj∈D. Then f(v)=∑djf(vj)=∑djg(j)=g(v)∈Ui⁢α for some i. This shows that f⁢(Ki⁢α)⊆Ui⁢α and we have f∈⋂B⁢(Ki⁢α,Ui⁢α)⊆B.

Conversely, assume that the ring R is a dense subset of the space EndD⁡(V). Let v1,…,vn be linearly independent, and w1,…,wn be arbitrary vectors in V. Let W be the subspacespanned by the vi’s. Because the vi’s are linearly independent, there exists a linear transformation g such that g⁢(vi)=wi and g⁢(v)=0 for v∉W. Let Ki={vi} and Ui={wi}. Then the Ki’s are compact and the Ui’s are open in the discrete space V. Clearly g∈{h∣h⁢(vi)=wi}=B⁢(Ki,Ui) for each i=1,…,n. So g lies in the neighborhood B=∩B⁢(Ki,Ui)⊆EndD⁡(V). Since R is dense in EndD⁡(V), there is an f∈R∩B. This implies that f⁢(vi)=wi for all i.
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Remarks.

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If V is finite dimensional over D, then any dense ring of linear transformations R=EndD⁡(V). This can be easily observed by using the second half of the proof above. Take a basis v1,…,vn of V and any set of n vectors w1,…,wn in V. Let g be the linear transformation that maps vi to wi. The above proof shows that there is an f∈R such that f agrees with g on the basis elements. But then they must agree on all of V as a result, which is precisely the statement that g=f∈R.